Question

If the acceleration due to gravity is g \[m{s^{ - 2}}\], a sphere of lead of density \[\delta {\text{ }}kg{m^{ - 3}}\] is gently released in a column of liquid of density d \[kg{m^{ - 3}}\] \[(\delta > d)\], the sphere will fall vertically with:
(A) An acceleration of 8 \[m{s^{ - 2}}\]
(B) No acceleration
(C) An acceleration of $g\left[ {1 - \dfrac{d}{\delta }} \right]m{s^{ - 2}}$
(D) An acceleration $g\left( {\dfrac{\delta }{d}} \right)m{s^{ - 2}}$

Answer 1

Hint There are two forces acting on the sphere. So, the net force on the sphere is calculated by taking the difference between the gravitational and buoyancy force. Then by using the relation \[F = {\text{ }}ma\] the acceleration can be evaluated.

  Complete step-by-step solution
When the sphere is falling through the column let its volume be V. Since, density is the ratio of mass to volume, mass of the sphere can be written as
\[m{\text{ }} = V{\text{ }}\delta {\text{ }}kg\]
When it is released in the liquid, the volume remains the same and the density is d. So, m inside the liquid can be written as \[m{\text{ }} = Vd{\text{ }}kg\].

During its motion down the column,
Gravitational force acting downwards is given by,
$
  {F_g} = mg \\
  {F_g} = V\delta g \\
 $
Buoyancy force acting upwards is given by,
$
  {F_b} = mg \\
  {F_g} = Vdg \\
 $
The net force acting is
$
  {F_{net}} = {F_g} - {F_b} \\
  {F_{net}} = Vg(\delta - d) \\
 $
Since acceleration is
$
  a = \dfrac{{{F_{net}}}}{m} \\
  a = \dfrac{{Vg(\delta - d)}}{{V\delta }} \\
  a = g\left( {1 - \dfrac{d}{\delta }} \right)m{s^{ - 2}} \\
 $
Hence the acceleration during its fall is $g\left( {1 - \dfrac{d}{\delta }} \right)m{s^{ - 2}}$
The correct option is C.

Note According to Archimedes principle, when a body is immersed partially or wholly in a fluid, at rest, it is buoyed up with force equal to the weight of the fluid displaced \[\left( {{w_{fluid}}} \right)\] by the body.
Also, the ratio of weight of the object immersed \[\left( {{w_{object}}} \right)\] to weight of displaced fluid \[\left( {{w_{fluid}}} \right)\] is equal to the ratio of the density of the object \[(\delta )\] to that of fluid (d)
$\left( {\dfrac{\delta }{d}} \right) = \left( {\dfrac{{{w_{onject}}}}{{{w_{fluid}}}}} \right)$